Latest TCS Aptitude Question SOLUTION: ab+a+b=250 bc+b+c=300 ac+a+c=216 then find a+b+c

a+b+c=41
a+b+ab=135 subject a i.e, a=(135-b)/(1+b)

a+c+ac=151 similarly subject c i.e, c=(151-a)/(1+a)
now replacing a=(135-b)/(1+b)
we get c=(2+19b)/17

b+c+bc=322 or,c=(322-b)/(1+b)

so we can write ((322-b)/(1+b))=((2+19b)/17 or,b^2+2b-288=0 or,b=either -18 or 16

now if b=-18 then a=9 & c=20;
if b=16 then a=7 & c=18

I don't know what's the use of asking these types of questions in an aptitude test...These are questions on functional inequality analysis....
Well, This problem can be solved in two methods (you are free to choose whichever you like...)
Note:Firstly, these type of equations don't have any exact solutions. (Sadly,this one is also not one of those exceptional types!!)Hence, solution to this equation are only approximates...

Method 1 :(The Conventional Way!!)
ab + a + b = 250
a(1+b) + b = 250
a = (250−b)/(1+b)

ac + a + c = 300
c(1+a) + a = 300
c = (300−a)/(1+a)=(300−(250−b)/(1+b))/(1+(250−b)/(1+b))
= (300(1+b)−(250−b))/1+b+250−b
= (300(1+b)−(250−b))/251
=(50+301b)/251
=(50+301b)/251

bc + b + c = 216
((50+301b)b/251)+b+((50+301b)/251)=216
(50 + 301b)b + 251b + 50 + 301b = 54216
301b^2 +602b−54216 =0
b=12.46

a =(250−b)/(1+b)=(250−12.46)/(1+12.46) = 17.65
c = (300−a)(1+a)=(300−17.65)/(1+17.65) = 15.14

These values of a,b and c statisfies the given conditions,as given below
ab + a + b = 17.65 *12.46 + 17.65 + 12.46 ≈ 250
ac + a + c = 17.65 *15.14 + 17.65 + 15.14 ≈ 300.21
bc + b + c = 12.46 *15.14 + 12.46 + 15.14 ≈ 216

Hence,
a + b + c = 17.65 + 12.46 + 15.14 = 45.25 [This may be one solution!! But remember there are many approximations which I have done here!!)

Method 2: (Inequality approximation method)
Take the three equations,
ab+a+b=250.....(1)
bc+b+c=300.....(2)
ac+a+c=216.....(3)

add them up, we get,
2(a+b+c)+ab+ac+bc=766....(4)
a+b+c= (766-(ab+ac+bc))/2....(5)

So, we are getting some form here,but once again solving this equation is hard as we don't know the value of ab,bc,ac here.

So, apply A.M.-G.M. Inequality (If you don't know..read them up in wikipedia!!)
Take equation (1),apply AM-GM
(ab+a+b)/3 >=(ab)^(2/3)
put the value of ab+a+b=250.
so, ab =< (+(250/3)^(1/2))^3 or (-(250/3)^(1/2))^3.........(6)
Similarly for equation 2 and 3 we see that 300 and 216 are both divisible by 3,so we get whole numbers!!
then,
bc =< (+(100)^(1/2))^3 or (-(100)^(1/2))^3......(7)
ac =< (+(72)^(1/2))^3 or (-(72)^(1/2))^3......(8)

Using Equation 5 and 6,7,8. Plug in combination of values,you get the solutions for a+b+c.

Please note: This kind of questions are purely options dependent. Try to nullify options.The second method is way faster than the first one...(according to me atleast!! but you are free to choose whatever you like!!)

I would appreciate if anyone comes up with a more novel solution to this problem..

ans-165(approx)

b=(300-c)/(c+1) & a=(216-c)/(c+1);now put these two values of a and b in eq^n 1 (above most),after that find the value of c from that eq^n that is c=15.131518,now find the valu of b and c with the help of value of c,finally the value of a=16 approx,value of b=12 approx and c=17 approx so the value of a+b+c=45 approx.......

can any one clearly explain

ab+a+b=250
Adding 1 on both sides
(a+1)(b+1)=251 ...(1)
Similarly
(b+1)(c+1)=301 ...(2)
(c+1)(a+1)=217 ...(3)
Multiply the 3 eqns. and take square root to get the value of(a+1)(b+1)(c+1) then divide by (1),(2),(3)to get (c+1),(a+1),(b+1)respectively.Then add and subtract 3

a+b+c=46...is the answer...

how ROHIT...???

HOW ROHIT ?

a+b+c will b near about 41...